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Question

A small block of mass m is placed at rest on the top of a smooth wedge of mass M,
which in turn is placed at rest on a smooth horizontal surface as shown in figure. Let h
be the height of wedge and θ is the inclination, then the distance moved by the wedge as
the block reaches the foot of the wedge is :

117692.PNG
  1. MhcotθM+m
  2. mhcotθM+m
  3. MhcosθM+m
  4. mhcosθM+m

A
MhcosθM+m
B
MhcotθM+m
C
mhcotθM+m
D
mhcosθM+m
Solution
Verified by Toppr

As there is zero external force in horizontal direction the COM(Centre of mass) will remain at same position.

The relative distance travelled by the block in right direction is x=hcotθ

Let the incline move by distance x in left direction

The net distance covered by box is hcotθx

Using m1x1=m2x2

m(hcotθx)=Mx

x=mhcotθm+M

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